/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 CALC The gravitational pull of t... [FREE SOLUTION] | 91Ó°ÊÓ

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CALC The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-\mathrm{kg}\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

Short Answer

Expert verified
The asteroid's minimum impact speed is approximately 11,200 m/s, and it imparts about 1.26 x 10^12 J of kinetic energy to Earth.

Step by step solution

01

Understand the Gravitational Force

The gravitational force between two masses is given by the formula: \[ F = \frac{G m_1 m_2}{r^2} \]where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses involved, and \(r\) is the distance between their centers. As the object approaches from a great distance, we can assume it starts with gravitational potential energy which converts to kinetic energy as it falls.
02

Set Up Conservation of Energy

The potential energy at a great distance is nearly zero and the asteroid has kinetic energy, which is zero. As it reaches the Earth's surface, all potential energy is converted to kinetic energy. The expression for energy conservation is: \[ 0 = -\frac{G M m}{R} + \frac{1}{2} m v^2 \]where \(M\) is Earth's mass, \(m\) is the asteroid's mass, \(R\) is the Earth's radius, and \(v\) is the velocity at impact.
03

Solve for Minimum Speed

Using the conservation of energy equation, solve for velocity \(v\):\[ v = \sqrt{\frac{2 G M}{R}} \]Plug in known values:\[ G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}, \, M = 5.972 \times 10^{24} \text{ kg}, \, R = 6.371 \times 10^6 \text{ m} \]. Calculate to find the minimum speed \(v\).
04

Calculate Kinetic Energy

Once the speed \(v\) is found, calculate kinetic energy using:\[ KE = \frac{1}{2} m v^2 \]where \(m = 20,000 \text{ kg}\). Substitute the speed value solved in Step 3 to find the kinetic energy imparted to Earth.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In the context of a falling asteroid, this principle helps us understand how potential energy is converted into kinetic energy as it moves closer to Earth. At a large distance from the Earth, the asteroid effectively has gravitational potential energy but almost no kinetic energy since it is relatively stationary. As the asteroid falls toward the Earth, this potential energy is converted into kinetic energy due to the gravitational pull of the planet.
This transformation is governed by the conservation of energy equation, which, in this case, looks like:\[0 = - \frac{G M m}{R} + \frac{1}{2} m v^2\]In this equation:
  • \(G\) is the gravitational constant.
  • \(M\) is the mass of the Earth.
  • \(m\) is the mass of the asteroid.
  • \(R\) is the radius of the Earth.
  • \(v\) is the velocity of the asteroid upon impact.
As the asteroid reaches Earth's surface, all of its potential energy is converted into kinetic energy, thus allowing us to calculate the speed at which it hits the planet.
Gravitational Force Equation
The gravitational force between two objects is described by Newton's law of universal gravitation. It states that every point mass attracts every other point mass by a force acting along the straight line joining the two points. The size of the force is proportional to the product of their two masses and inversely proportional to the square of the distance between their centers. Mathematically, this is expressed as:\[F = \frac{G m_1 m_2}{r^2}\]Here,
  • \(F\) is the gravitational force.
  • \(G\) is the gravitational constant \(6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\).
  • \(m_1\) and \(m_2\) are the masses of the two objects, in this case, the Earth and the asteroid.
  • \(r\) is the distance between the centers of the two masses.
As the asteroid falls towards Earth, this force is what accelerates it, causing its potential energy to convert into kinetic energy.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When the asteroid is falling towards the Earth, it accelerates and gains speed. This increase in speed is directly related to an increase in kinetic energy, which can be calculated using the formula:\[KE = \frac{1}{2} m v^2\]where
  • \(KE\) is the kinetic energy.
  • \(m\) is the mass of the object, which in our case, is the 20,000 kg asteroid.
  • \(v\) is the velocity of the object upon impact.
In the scenario described, once we calculate the minimum speed \(v\) of the asteroid when it strikes the Earth, we can easily determine the amount of kinetic energy transferred to the planet. This information can help us understand both the potential damage an object of this size could cause and the impressive nature of energy transformation processes guided by physical principles like the conservation of energy.

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Most popular questions from this chapter

A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal roadway using a cable having a tension of 850 \(\mathrm{N}\) . (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

You and your bicycle have combined mass 80.0 \(\mathrm{kg} .\) When you reach the bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\) Fig. \(\mathrm{P} 6.78)\) . At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s} .\) You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

A student proposes a design for an automobile crash barrier in which a 1700 -kg sport utility vehicle moving at 20.0 \(\mathrm{m} / \mathrm{s}\) crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00\(g .\) (a) Find the required spring constant \(k,\) and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

A 75.0 -kg painter climbs a ladder that is 2.75 \(\mathrm{m}\) long leaning against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\) -kg automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizon- tally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h}\) ?

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